Let \(f\colon G={2}{k}\rightarrow H\)  be a continuous homomorphism into a topological group \(H\) .
By [1]}, any non-trivial normal subgroup of \(G\)  is co-compact, thus if \(f\)  is not injective \(f(G)\)  is compact and obviously closed.
We thus assume as we may that \(f\)  is injective.
By Proposition REF , the \(f\) -initial topology \(\tau ^f\)  must coincide with the analytic topology on \(G\) .
Therefore, we get that \(f\)  is a homeomorphism onto its image.
Therefore the subspace topology on \(f(G)\)  is Polish, hence closed, see [2]}.
